Question: Evaluate $\int^\pi_0\sin^2x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12\pi\\$ (Choice B) B $\pi$ (Choice C) C $\dfrac32\pi\\$ (Choice D) D $2\pi$
Explanation: Since the integrand has only an even power of $~\sin x\,$, we use the identity $\cos2x=1-2\sin^2x\,$. $ \int^\pi_0\sin^2x\, dx = \dfrac12\int^\pi_0(1-\cos2x)\,dx$ We continue with the antiderivative. $ \int^\pi_0\sin^2x\, dx = \dfrac12\int^\pi_0(1-\cos2x)\,dx$ $ ~~~~~ =\dfrac12\Big(x-\dfrac12\sin2x\Big)\Bigg|^\pi_0$ $ ~~~~~=\dfrac12\Big(\pi-\dfrac12\sin2\pi\Big)-\dfrac12\Big(0-\dfrac12\sin0\Big)$ $ ~~~~=\dfrac12\pi$